C. Bear and String Distancetime limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.
The distance between two letters is defined as the difference between their positions in the alphabet. For example, 
, and 
.
Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, 
, and 
.
Limak gives you a nice string s and an integer k. He challenges you to find any nice string s‘ that 
. Find any s‘ satisfying the given conditions, or print “-1” if it’s impossible to do so.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usegets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.
Input
The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).
The second line contains a string s of length n, consisting of lowercase English letters.
Output
If there is no string satisfying the given conditions then print “-1” (without the quotes).
Otherwise, print any nice string s‘ that .
Examplesinput
4 26
bear
output
roar
input
2 7
af
output
db
input
3 1000
hey
output
-1
题意：
AC代码：
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
char s[N];
int n,k;
int main()
{
int num=0;
scanf("%d%d",&n,&k);
scanf("%s",s);
for(int i=0;i<n;i++)
{
num+=max(s[i]-'a','z'-s[i]);
}
if(k>num)cout<<"-1"<<"\n";
else
{
int sum;
for(int i=0;i<n;i++)
{
sum=max(s[i]-'a','z'-s[i]);
//cout<<sum<<endl;
if(sum<=k){
k-=sum;
if(s[i]-'a'>'z'-s[i])s[i]='a';
else s[i]='z';}
else if(sum>k&&k!=0)
{
if(s[i]-'a'>=k)
{
s[i]=s[i]-k;
}
else
{
s[i]=s[i]+k;
}
k=0;
}
printf("%c",s[i]);
} }
return 0;
}
